Wait, why do we use the discriminant formula $|\alpha - \beta| = \frac{\sqrt{D}}{|a|}$? Can't we just use the quadratic formula?

You absolutely can! The quadratic formula gives you the roots as $\frac{-b \pm \sqrt{D}}{2a}$. If you subtract these two roots, you get $\frac{\sqrt{D}}{a} - (\frac{-\sqrt{D}}{a}) = \frac{2\sqrt{D}}{2a} = \frac{\sqrt{D}}{a}$. Using the difference formula is just a faster, more elegant way to handle the 'consecutive' condition without carrying the full root expressions.

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Animated Solution for Mathematics - Quadratic Equations: If, for a positive integer n, the quadratic equation, $x (x + 1) + (x + 1) (x + 2) + \dots + (x + n - 1) (x + n) = 10 n$ has two consecutive integral solutions, then n is equal to:

Q: Why did we divide the entire equation by $n$ in the middle of the process?

That is a great observation! We are given that $n$ is a positive integer. In mathematics, we can only divide by an expression if we are certain it is not zero. Since $n \geq 1$, we know $n \neq 0$, making the division perfectly valid and simplifying our quadratic equation significantly.

Q: I'm stuck on the summation of $(x + k - 1)(x + k)$. How do I handle the $x$ terms?

Remember that when you are summing with respect to $k$, any term that does not contain $k$ is treated as a constant. So, $\sum_{k=1}^{n} x^2$ is just $n \cdot x^2$. Treat $x$ as a constant, and you will find the algebra becomes much cleaner.

Select Answer:

Visualized Solution

Analyzing the Given Equation

Given: $x (x + 1) + (x + 1) (x + 2) + \dots + (x + n - 1) (x + n) = 10 n$
Condition: The equation has two consecutive integral solutions.
Goal: Find the value of the positive integer $n$ .

Compressing with $Σ$ Notation

Represent the sum using sigma notation:
$\sum_{k = 1}^{n} (x + k - 1) (x + k) = 10 n$

Expanding the General Term

Expand the general term $(x + k - 1) (x + k)$ :
$= x^{2} + (k + k - 1) x + k (k - 1)$
$= x^{2} + (2 k - 1) x + k^{2} - k$

Distributing the Summation

Distribute $\sum$ over the expanded terms:
$\sum_{k = 1}^{n} x^{2} + x \sum_{k = 1}^{n} (2 k - 1) + \sum_{k = 1}^{n} (k^{2} - k) = 10 n$

Standard Summation Formulas

Recall standard formulas:
$\sum_{k = 1}^{n} 1 = n$
$\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$
$\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$\sum_{k = 1}^{n} (2 k - 1) = n^{2}$

Substituting the Formulas

Substitute into the equation:
$n x^{2} + x (n^{2}) + [\frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2}] = 10 n$

Simplifying the Constant Term

Simplify $\frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2}$ :
Take $\frac{n ( n + 1 )}{2}$ common:
$= \frac{n ( n + 1 )}{2} [\frac{2 n + 1}{3} - 1]$
$= \frac{n ( n + 1 )}{2} [\frac{2 n - 2}{3}] = \frac{n ( n ^{2} - 1 )}{3}$

Standard Quadratic Form

The equation becomes: $n x^{2} + n^{2} x + \frac{n ( n ^{2} - 1 )}{3} = 10 n$
Divide by $n$ (since $n > 0$ ):
$x^{2} + n x + \frac{n ^{2} - 1}{3} = 10$
Rearrange: $x^{2} + n x + \frac{n ^{2} - 31}{3} = 0$

Condition for Consecutive Roots

Let the roots be $α$ and $β$ .
Given: Roots are consecutive integers.
Therefore, the difference of roots is $1$ : $∣ α - β ∣ = 1$

Difference of Roots Formula

Formula: $∣ α - β ∣ = \frac{D}{∣ a ∣}$
Here, $a = 1$ , so $D = 1 ⟹ D = 1$
Discriminant $D = b^{2} - 4 a c$

Setting up the Discriminant

Substitute $a = 1$ , $b = n$ , $c = \frac{n ^{2} - 31}{3}$ into $D = 1$ :
$n^{2} - 4 (1) (\frac{n ^{2} - 31}{3}) = 1$

Solving for $n$

Multiply by $3$ : $3 n^{2} - 4 (n^{2} - 31) = 3$
Expand: $3 n^{2} - 4 n^{2} + 124 = 3$
Simplify: $- n^{2} + 124 = 3 ⟹ n^{2} = 121$
Since $n$ is a positive integer: $n = 11$

00:00 / 00:00

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The equation

x (x + 1) + (x + 1) (x + 2) + \dots + (x + n - 1) (x + n) = 10 n

may appear intimidating, but it is a structured sequence. By using Sigma notation, we can compress this expression into a manageable entity:

k = 1 \sum n (x + k - 1) (x + k) = 10 n

By identifying the general term, we have already won half the battle. This is the language of efficiency in JEE Advanced mathematics.

The Algebraic Grind

Now, let us expand the general term

(x + k - 1) (x + k)

. Treating

x

as a constant relative to the summation index

k

, the expansion becomes

x^{2} + (2 k - 1) x + k^{2} - k

Distributing the summation operator across these terms yields:

n x^{2} + x k = 1 \sum n (2 k - 1) + k = 1 \sum n (k^{2} - k) = 10 n

We now apply the standard summation formulas:

\sum_{k = 1}^{n} 1 = n

\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}

, and

\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

The Quadratic Reveal

Substituting these formulas into our equation and simplifying the terms, we arrive at a clean, standard quadratic equation:

x^{2} + n x + \frac{n ^{2} - 31}{3} = 0

This is the pivot point of the problem. We have reduced a complex series into a simple quadratic equation where the coefficients are functions of

n

The Geometric Insight

We are given that the roots are consecutive integers. Geometrically, this means the distance between the roots

α

and

β

on the number line is exactly

1

, or

∣ α - β ∣ = 1

The difference of roots for a quadratic

a x^{2} + b x + c = 0

is given by

\frac{D}{∣ a ∣}

. Since our leading coefficient

a = 1

, this simplifies to

D = 1

, or

D = 1

Setting the discriminant

b^{2} - 4 a c = 1

, we obtain:

n^{2} - 4 (1) (\frac{n ^{2} - 31}{3}) = 1

The Final Victory

To solve for

n

, we first multiply by

3

to clear the fraction:

3 n^{2} - 4 (n^{2} - 31) = 3

3 n^{2} - 4 n^{2} + 124 = 3

- n^{2} = - 121 \Rightarrow n^{2} = 121

Since

n

must be a positive integer, we conclude that $n = 11$ . What started as a terrifying series ended as an elegant, logical progression.

Common Traps & FAQs

Wait, why do we use the discriminant formula

∣ α - β ∣ = \frac{D}{∣ a ∣}

? Can't we just use the quadratic formula?

You absolutely can! The quadratic formula gives you the roots as

\frac{- b \pm D}{2 a}

. If you subtract these two roots, you get

\frac{D}{a} - (\frac{- D}{a}) = \frac{2 D}{2 a} = \frac{D}{a}

. Using the difference formula is just a faster, more elegant way to handle the 'consecutive' condition without carrying the full root expressions.

Why did we divide the entire equation by

n

in the middle of the process?

That is a great observation! We are given that

n

is a positive integer. In mathematics, we can only divide by an expression if we are certain it is not zero. Since

n \geq 1

, we know

n \neq = 0

, making the division perfectly valid and simplifying our quadratic equation significantly.

I'm stuck on the summation of

(x + k - 1) (x + k)

. How do I handle the

x

terms?

Remember that when you are summing with respect to

k

, any term that does not contain

k

is treated as a constant. So,

\sum_{k = 1}^{n} x^{2}

is just

n \cdot x^{2}

. Treat

x

as a constant, and you will find the algebra becomes much cleaner.

Animated Solution for Mathematics - Quadratic Equations: If, for a positive integer n, the quadratic equation, $x (x + 1) + (x + 1) (x + 2) + \dots + (x + n - 1) (x + n) = 10 n$ has two consecutive integral solutions, then n is equal to:

Select Answer:

Visualized Solution

Analyzing the Given Equation

Compressing with $Σ$ Notation

Expanding the General Term

Distributing the Summation

Standard Summation Formulas

Substituting the Formulas

Simplifying the Constant Term

Standard Quadratic Form

Condition for Consecutive Roots

Difference of Roots Formula

Setting up the Discriminant

Solving for $n$

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Algebraic Grind

The Quadratic Reveal

The Geometric Insight

The Final Victory

Similar Questions

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Comprehension Passage

$a_{12} =$

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Select Answer:

Visualized Solution

Analyzing the Given Equation

Compressing with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Σ Notation

Expanding the General Term

Distributing the Summation

Standard Summation Formulas

Substituting the Formulas

Simplifying the Constant Term

Standard Quadratic Form

Condition for Consecutive Roots

Difference of Roots Formula

Setting up the Discriminant

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Algebraic Grind

The Quadratic Reveal

The Geometric Insight

The Final Victory

Similar Questions

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Comprehension Passage

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